Question #5edfb

2 Answers
Dec 7, 2017

#10923#
See explanation.

Explanation:

So you have a sequence that goes:
3, 8, 13, 18,...

One possibility that I see right away is that #8-3 = 5#
then #13-8 =5#, then #18-13=5#, so perhaps the sequence is just adding 5 to the previous number, starting from 3.

If we call the first entry, #3=k_1#,
and the second entry of sequence, #8=k_2#,
we can rewrite the first and second entries as follows:
#k_1=3#
#k_2=k_1+5#
the third entry would be written:
#k_3=k_2+5#
fourth entry:
#k_4=k_3+5#
...
sixty-sixth entry:
#k_66=k_65+5#

We want the sum of #k_1# to #k_66#, that is, the sum of all the equations shown above.
We now notice that we can rewrite these equations as such:
#k_1=3#
#k_2=3+5#
#k_3=3+5+5#
#k_4=3+5+5+5#
...
#k_i=3+(i-1)*5#

Thus the sum can be written:
#\Sigma_(i=1)^{66} [3+(i-1)*5]#
which can be simplified to:
#\Sigma_(i=1)^{66} 3 + 5*\Sigma_(i=1)^{66} [(i-1)]#
and even to:
#\Sigma_(i=1)^{66} 3 + 5*[\Sigma_(i=1)^{66} i - \Sigma_(i=1)^{66}1]#
so:
#66*3 + 5*[67*66/2 - 66]#
#=10923#

Dec 7, 2017

#S_(66)=10923#

Explanation:

#"the sum to n terms of an arithmetic sequence is"#

#•color(white)(x)S_n=n/2[2a+(n-1)d]#

#"where a is the first term and d the common difference"#

#"here "a=3" and "d=18-13=13-8=8-3=5#

#rArrS_(66)=33[(2xx3)+(65xx5)]#

#color(white)(rArrS_(66))=33(6+325)#

#color(white)(rArrS_(66))=10923#